Since the underlying algebraic groups are just the usual free groups the injection is immediate. It is not too hard to see that the subgroup $\langle A\rangle $ generated by $A$ in $F_{M}(X)$ is closed (this shows up in the paper "The free topological group over the rationals"). But what about the embedding part? Does this always hold? If not, what if $X$ is a metric space? If it can be helped, I would like to avoid placing compactness conditions on $A$ and $X$.

Update: I have recently come upon a result which makes this easier. A theorem of O.V. Sipacheva in Free topological groups of spaces and their subspaces says that for Tychonoff $X$ and arbitrary $A\subset X$, $i:F_{M}(A)\rightarrow F_{M}(X)$ is a topological embedding if and only if $A$ is P-embedded in $X$. This is also equivalent to saying that $A$ with its universal uniformity is a uniform subspace of $X$ with its universal uniformity.

So I guess the updated question is: If $A$ is closed in Tychonoff $X$, then does $A$ with its universal uniformity become a uniform subspace of $X$ with its universal uniformity. If not, what are some sufficient conditions (on $A$) for this to occur?